This lecture
(September 19, 2007)
continues our first introduction to knot theory.
We will treat
the following notions.
A
braid
on strands
is a collection of curves in Euclidean space
starting at the
points
and ending at the points
for
, each meeting the planes
,
with
,
in a unique point.
In particular, the curves, referred to as strands, are disjoint and
descend monotonically.
Just like knots and links, braids are considered equal when
they are isotopic. At each instance of a smooth deformation realizing
the isotopy, the braid should remain a braid; in particular, the end
points of the braid remain fixed.
Just like knots and links, we will depict braids by means of diagrams;
in this case we take the projections to be orthogonal onto a vertical plane
behind the braid. Crossings are treated as for links, with overpasses and
underpasses. Because of our restrictions on braids, no Reidemeister I move
will be realizable. Reidemeister II and III moves are possible. For instance,
in the above diagram, a sequence of two Reidemeister II moves shows that the
corresponding braid is (isotopic to) the braid pictured below.
The strands in the braid are numbered 1,..., from left
to right, so that the leftmost strand is always strand 1 (no matter
what position it started in).
The
elementary braid
is the braid interchanging the two strands numbered
and
,
where the former underpasses the latter and
all other strands go straight from top to bottom.
The
inverse of the elementary braid
is the braid,
denoted
,
that interchanges the two strands numbered
and
,
where the former overpasses the latter and
all other strands go straight from top to bottom.
We will soon see why it makes sense to speak of an inverse here.
Two braids, say and ,
on strands can be composed
into a single braid, denoted
,
by placing under , thereby identifying the top ends of
with the bottom ends of , then viewing these identified
points as ordinary points on the resulting longer strands,
and scaling so that
the top ends of stay in place and the bottom ends of
are placed at
.
This composition turns the set of all braids on strands into
a group,
denoted
.
For, the composition is clearly associative, the braid all of whose
strands go straight down is the identity element with respect to the
composition, and the inverse of a braid is obtained by reflecting its
diagram in the horizontal line
at the bottom of the diagram (and moving the diagram by a translation so as to
put the end points back to their standard
positions). The latter follows from applications of Reidemeister II moves
to find an isotopy from the product of the braid and its inverse to
the trivial braid.
The group inverse of the elementary braid
now coincides with
,
as defined in the previous section; this explains the terminology chosen there.
Every braid is (isotopic to) the
product of elementary braids.
The following result by Emil Artin even
characterizes when two diagrams represent the same braid.
The braid group
on strands
has the following presentation on the generators
with
.
when
,
and
when
.
It is easy to deduce that the relations should hold;
see the two pictures below.
It is harder to see that these relations give a presentation, that is, that
every relation is a consequence of these. This was first proved by Emil Artin.
QED
As a consequence of the homogeneity of the defining relations
for
, we can define a homomorphism of groups
:
by decreeing
.
This implies that we can talk about the
height of a
braid
.
(Its absolute value
is a lower bound on the number of crossings needed to
draw it.)
The
writhe
of the closure of a braid coincides with the
height of
the braid.
A theorem by Alexander
shows that every oriented link is
equivalent to one obtained by identifying starting and ending
points above each other in a suitable braid.
Let be a braid. The
closure
of
is the oriented link obtained by connecting each endpoint
with its vertical counterpart
by means of an unknotted straight strand that does not cross any of
the other strands. The result is a link. Orienting all strands
of downward, we find an oriented link.
The closure of the identity element of
is the disentangled union of unknots.
This shows that it is important to specify the number of strands of
the braid.
The closure of a braid is a uniquely determined oriented link.
Different braids may represent the same oriented link.
Let be a braid.
We distinguish the following two kinds of Markov move.
-
Conjugation: If
, replace by
.
-
Embedding:
if
, replace by
or by
.
The reverses of these moves are also counted as Markov moves.
Two braids represent the same oriented link if and only if one can be obtained
from the other by a sequence of Markov moves.
Alexander
realized that every oriented knot can be written
as the closure of a braid. We represent an algorithm that, when given an
oriented link diagram, changes the diagram without changing the corresponding
oriented link, so that it becomes the closure of a braid.
This algorithm is due to Vogel.
The major steps in the algorithm are
-
Draw the Seifert circles.
-
Spot a pair of
Seifert circles whose orientations do not match in the sense that there is a
region without strands in whose boundary appear two strands of different
Seifert circles with non parallel directions.
-
Resolve this conflict by
applying Reidemeister II so as to let one of the two strands undercross the
other twice.
-
It may be convenient to make Seifert circles without conflicts concentric;
apply, for
this purpose,
Reidemeister 0
to a strand of a Seifert circle at the outside of the diagram.
-
Repeat this operation sufficiently many times in order to remove all
conflicts.
-
Cut the Seifert circles open in such a way that the braid appears.
HOMEWORK CHALLENGE 2 (3 pts): Show termination of this algorithm.
Here is an example.
The resulting braid in the example is
.
The Vogel algorithm, live in GAP:
-
-
-
Another example of the application of Vogel moves is shown in
the sequence of pictures below:
The implementation of this algorithm by Dan Roozemond is also
available online
.
Seifert surfaces can be visualized with
Seifertview
, by
Jack van Wijk
.
The above braid is represented in Seifertview by
ABCBBaBcB.
The Burau representation of the braid group
on strands is determined by the following images of its generators.
The -th generator
of
is mapped to the square matrix of size
of the following form: all of its rows and columns are as in the identity
matrix except for the -th and the
()-st; the square submatrix of size 2
with entries in
is
The map above, given on the generators of the Braid group
on strands, determines a unique linear representation
:
.
Let be the linear functional on
obtained by taking the standard inner product of a vector with
.
Then the submodule
of all vectors in
such that
is invariant under the action
of
.
The representation
is called the
Burau
representation
of . The restriction of
to is called the
reduced Burau
representation
of , and denoted
.
As a consequence, the map sending a braid to the characteristic
polynomial of its image under
, or any of its coefficients, is
invariant under conjugacy, and hence under one of the two Markov moves. By
suitably scaling the traces for varying we can accommodate for the
other Markov move and obtain link
invariants.
We apply this principle to the characteristic polynomial of the
representation on .
Note that is a free module
over
with basis
the column vectors
.
Let be a link.
The
Alexander polynomial
of a braid on
strands is the Laurent polynomial
,
where
is the reduced Burau representation of
.
The Alexander polynomial is the same on each braid whose closure
represents a fixed oriented link. In particular, it is an
oriented link invariant.
Example 1.
The trefoil is represented by
,
viewed as an element of
,
so
.
The action of
on the submodule of
, which
is spanned by
,
is multiplication by
,
and so
follows from
.
Conway observed that the Alexander polynomial can be computed by means of
a simple recursion. Consider the initial rule
if L is the trivial link with one component, and the following recursive rule:
,
where
are related as follows to an oriented link diagram in which
, say, is a crossing.
We let
be with replaced by two non-crossing strands.
If the sign of is positive,
then we take
and
we let
be with replaced by a negative crossing.
Similarly, if
the sign of is negative.
Note that, using the recursive rule appropriately, one can get
down from any link to a link with no crossings, but possibly
consisting of more than one component. Some additional applications of
the rule (in the other direction, creating new crossings) may be
required to finish the process.
Call the polynomial obtained from the above recursion the
Conway polynomial
of the oriented link .
The Conway polynomial of an oriented link
is equal to
up to a multiple by an invertible element of
.
Example 2.
The picture below shows the stages of a computation of the
Conway polynomial of the trefoil.
-
Braids (.5 pts)
Write the braid in the picture as a product of elementary braids and
their inverses.
-
The braid group (2 pts)
We exhibit elements in the center of the braid groups.
-
Show that the braid
satisfies
.
Conclude that
lies in the center of
.
-
Show that the braid
satisfies
.
Conclude that
lies in the center of
.
-
Draw the braid
.
-
Try and prove a statement analogous to the first two items for
braids with strands.
-
The braid group (1.5 pts)
By
we denote the symmetric group on letters.
Let :
be the map that sends the braid to the permutation
sending
to if the strand with end point
at the bottom of
ends at the point
at the top.
Prove that is a surjective homomorphism of groups.
Verify that the map can be described in terms of
braid diagrams as the
removal of the overpass and underpass information at each crossing.
-
The closure of a braid (1 pt)
Compare the two braids
and
. Show that the linking
numbers of the closures of the two braids differ.
Conclude that the braids
and
are not conjugate in
.
-
The closure of a braid (1 pt)
Express the reflection
and the reverse of the closure of a braid as the
closure of a braid expressed in terms of .
-
Markov moves (1 pt)
Show that the braid
can be transformed into
by a series of Markov moves.
-
Vogel algorithm (1 pt)
Apply the Vogel algorithm to the knot depicted below.
Compare the braid you obtain with the result of the interactive program
in the notes that produces a braid upon input a
signed Gauss code
of the knot.
-
Vogel algorithm (2 pts)
This is a more complicated version of the
previous exercise.
Apply the Vogel algorithm to the knot depicted below.
Compare the braid you obtain with the result of the interactive program
in the notes that produces a braid upon input a
signed Gauss code
of the knot.
-
The Burau representation (1 pt)
Prove the statement, made in the notes, that is a free module
over
with basis
the column vectors
.
-
The Burau representation (1 pt)
We express the height
of a braid in terms of the Burau representation.
Prove
.
-
The Burau representation (1 pt)
Show that the matrix of
on the basis of found
above
has all rows equal to those of the identity matrix except
for the -th, which is
with the entry
on the diagonal (and other entries truncated, at the extremes).
-
Alexander polynomial (3 pts)
This is a challenge problem. You are allowed to use computer algebra
packages to carry out the work. A a nontrivial knot is exhibited whose
Alexander polynomial does not separate it from the trivial knot. One of the
complications of this problem is that the knot is not given as the closure of
a braid.
Consider the braid
.
Connect horizontally (at the top as well as at the bottom of the diagram)
the endpoints 2 and 3, the endpoints 4 and 5,
and the endpoints 1 and 6, making sure that no new crossings are introduced.
Prove that the Alexander polynomial of the knot thus obtained
(and oriented in any way you like)
is trivial.
Prove that this knot is not trivial. Conclude that the Alexander
polynomial does not separate all knots from the unknot.
-
Conway polynomial (1.5 pts)
Compute the Conway polynomial of the knot depicted
in the diagram with six crossings above.
Compare the result with the Alexander polynomial obtained from the definition
by means of a determinant of a matrix involving the Burau representation.
(For this purpose, it makes sense to do first the exercise
to which the knot diagram
belongs.)